Explanation:
yooooooooo
what is the force of gravitational attraction between a 92 kg student and a 550 g slice of pizza that are 25 cm apart
Answer:
[tex]F = 5.4*10^{-8}\ N[/tex]
Explanation:
Given
Represent the mass of the student with M and the mass of the slice of pizza with m
[tex]M = 92kg[/tex]
[tex]m = 550g[/tex]
[tex]d = 25cm[/tex]
Required
Determine the force of attraction
This is calculated as:
[tex]F = \frac{GMm}{d^2}[/tex]
Where G = gravitational constant
[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]
Convert both mass to kilogram and distance to metre
[tex]m = 550g[/tex]
[tex]m = 550kg/1000[/tex]
[tex]m = 0.55kg[/tex]
[tex]d = 25cm[/tex]
[tex]d = 25m/100[/tex]
[tex]d = 0.25m[/tex]
Substitute these values in [tex]F = \frac{GMm}{d^2}[/tex]
[tex]F = \frac{6.67408 * 10^{-11} * 92 * 0.55}{0.25^2}[/tex]
[tex]F = \frac{6.67408 * 92 * 0.55* 10^{-11} }{0.25^2}[/tex]
[tex]F = \frac{337.708448* 10^{-11} }{0.0625}[/tex]
[tex]F = 5403.335168* 10^{-11}[/tex]
[tex]F = 5.403335168* 10^3*10^{-11}[/tex]
[tex]F = 5.403335168*10^{3-11}[/tex]
[tex]F = 5.403335168*10^{-8}[/tex]
[tex]F = 5.4*10^{-8}\ N[/tex]
explain what the emitter does in CT
the emitter of x-rays rotates around the patient and the detector, placed in diametrically opposite side, picks up the image of a body section (beam and detector move in synchrony)
i believe that should be ur answer =) good luck !
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?
A 0.31 s
B 0.56 s
C 4.3s
D 70s
Answer:
C. 4.3 seconds
Explanation:
B 0.56 s is the time period of a twirlers baton.
What is Centripetal Acceleration?Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.
Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.
The centripetal acceleration is given by:
a = 4π²R/T²
Given values are:
a = 47.8 m/s²
D = 0.76 m so , R = 0.76/2 = 0.38m
Using this formula,
47.8*T² = 4π² x0.38
T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]
T = 0.56 s
Therefore,
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.
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.
What happens to the electrical energy that does not become light energy?
A. The lightbulb transforms it into mechanical energy,
B. The lightbulb transforms it into thermal energy.
C. Some of the energy is destroyed over time rather than being
conserved
D. New energy is produced in the system when the lightbulb creates
light energy
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Answer:
B
Explanation:
To solve this we must be knowing each and every concept related to electrical energy. Therefore, the correct option is option B among all the given options.
What is electrical energy?The work done by an electric charge is referred to as electrical energy. For time t seconds, if electricity I ampere passes throughout a conductor or any other conducting element with a potential differential v volts across it.
The kilowatt hour is both a practical as well as an economic unit of electrical energy. The basic commercial unit is the watt-hour, one and kilowatt hour equals 1000 watt hours. Electric supply providers charge their customers every kilowatt hour unit of electricity used. This kilowatt hour seems to be a BOT unit, or board of trade unit. The electrical energy that does not become light energy, the lightbulb transforms it into thermal energy.
Therefore, the correct option is option B among all the given options.
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Help!!! Need answer ASAP.
Answer:
a = 11.03 [m/s²]
Explanation:
To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force = 160000 [N]
m = mass = 14500 [kg]
a = acceleration [m/s²]
160000 = 14500*a
a = 11.03 [m/s²]
Help!!! Need answer ASAP.
Answer:
a = 0.5 [m/s²]
Explanation:
To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.
ΣF = m*a
F = force = 200 [N]
m = mass = 400 [kg]
a = acceleration [m/s²]
Now replacing:
200 = 400*a
a = 0.5 [m/s²]
A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a perfectly elastic collision, what will be the velocity of the small ball if the heavier ball has a speed of 2m/s after collision.
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
What ate the two safety precautions that should be taken before driving your car?
Answer:
When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.
Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.
A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the horizontal. The ball rolls without slipping down the incline and at the bottom has a speed of 4.9 m/s. How many revolutions does the ball rotate through as it rolls down the incline
Answer:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
The ball rotates 6.78 revolutions.
Explanation:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
At the bottom the ball has the following angular speed:
[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]
Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:
[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]
To find the revolutions we need the time, which can be found using the following equation:
[tex] v_{f} = v_{0} + at [/tex]
[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)
So first, we need to find the acceleration:
[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]
Since it starts from rest (v₀ = 0):
[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]
Finally, we can find the revolutions:
[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]
Therefore, the ball rotates 6.78 revolutions.
I hope it helps you!
A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land?
Answer:
Explanation:
horizontal component of velocity of throw = 20 cos20 = 18.8 m /s
vertical downwards component = 20 sin20 = 6.84 m /s
time to displace by height 30 m = t , initial velocity u = 6.84 m /s
h = ut + 1/2 gt²
30 = 6.84 t + .5 x 9.8 t²
4.9 t² + 6.84 t - 30 = 0
t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9
= - 6.84 ±√( 46.78 + 588 ) / 9.8
= - 6.84 ±√(634.78 ) / 9.8
= - 6.84 ±25.2 / 9.8
= 1.87 s
horizontal displacement in 1.87 s
= 18.8 x 1.87
= 35.15 m .
June does an experiment to study how salt affects the freezing point of water
Answer:
Salt melts ice and helps keep water from re-freezing by lowering the freezing point of water. This phenomenon is called freezing point depression. Salt only helps if there is a little bit of liquid water available. The salt has to dissolve into its ions in order to work.
Explanation:
1. It plays a vital role for self- expression and has been part of rituals
and religious gatherings.
B. Dance
A. Aerobic activity
C. Fitness D. Zumba
Answer:
b.dance
Explanation:
don't know the explanation
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A.
Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy?
Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude?
Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3?
Answer:
a) The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) The maximum kinetic energy is increased by a factor of 9.
Explanation:
a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system ([tex]E[/tex]), measured in joules, is the sum of the translational kinetic energy ([tex]K[/tex]), measured in joules, and elastic potential energy ([tex]U[/tex]), measured in joules. That is:
[tex]E = K + U[/tex] (1)
By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
[tex]U = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (3)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity of the mass, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Elongation of the spring, measured in meters.
If we know that [tex]U = \frac{1}{9}\cdot E[/tex], [tex]k = k[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then:
[tex]\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}[/tex]
[tex]\frac{1}{9}\cdot A^{2} = x^{2}[/tex]
[tex]x= \frac{1}{3}\cdot A[/tex]
The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) If we know that [tex]k = k[/tex], [tex]x = \frac{1}{2}\cdot A[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then the equation of energy conservation associated with the system is:
[tex]\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K[/tex]
[tex]K = \frac{1}{4}\cdot k\cdot A^{2}[/tex]
The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.
A golf ball hit off a tee on level ground, lands 62 m away 3.0 later. What was the initial velocity of the golf ball?
62×3.0
think so not sure
Is a windmill that is not running, potential or kinetic?
Lifting a stone block 146m to the top of the Great Pyramid required 146,000 J of work. How much work was done to lift the block halfway to the top?
A. 36,500 Joules
B. 73,000 Joules
C. 146,000 Joules
D. 292,000 Joules
Please help me.
A 7300N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a turning cylindrical shaft.
If the shaft's diameter can be no larger than 12.0cm due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator
Answer:
The value is [tex]\alpha = 24.5 \ rad/s^2[/tex]
Explanation:
From the question we are told that
The weight of the elevator is [tex]W = 7300 \ N[/tex]
The acceleration it is to be given is [tex]a = 0.150 \ g = 0.150 * 9.8 = 1.47 \ m/s^2[/tex]
The diameter of the shaft is [tex]d = 12.0 \ cm = 0.12 \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.12}{2}[/tex]
=> [tex]r = 0.06 \ m[/tex]
Generally the minimum angular acceleration is mathematically represented as
[tex]\alpha = \frac{a}{r}[/tex]
=> [tex]\alpha = \frac{1.47}{0.06}[/tex]
=> [tex]\alpha = 24.5 \ rad/s^2[/tex]
How do we use energy transformation in our daily lives?
Answer:hat are some examples of energy transformation?
The Sun transforms nuclear energy into heat and light energy.
Our bodies convert chemical energy in our food into mechanical energy for us to move.
An electric fan transforms electrical energy into kinetic energy.
Explanation:
Which energy transformation occurs after a skydiver reaches terminal velocity? Gravitational potential energy transforms into thermal energy. Gravitational potential energy transforms into kinetic energy. Kinetic energy transforms into thermal energy. Kinetic energy transforms into gravitational potential energy. The answer is A. just took it
Answer:
a
Explanation:
The energy transformation occurs after a skydiver reaches terminal velocity is follows as;
A. Gravitational potential energy transforms into thermal energy.
B. Gravitational potential energy transforms into kinetic energy.
What is the gravitational potential energy?The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:
U = mgh
where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground.
The skydiver gravitational potential energy decreases as the altitude decreases and his kinetic energy store increases as his speed increases.
When a skydiver jumps out of a plane, the energy transfers take place as;
The skydiver's kinetic energy store increases as their speed increases and the thermal store of the air and the skydiver increases, as there is friction between the skydiver and the air particles.
In the given situation, both options A and B are correct.
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A box of mass 7.0 kg is accelerated from rest across a floor at a rate of 2.0 m/s2 for 9.0 s .Find the net work done on the box. Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
Step one:
given data
mass = 7kg
acceleration =2m/s^2
time= 9seconds
acceleration = velocity/time
velocity= acceleration *time
velocity=2*9
velocity= 18m/s
distance moved= velocity* time
distance= 18*9
distance=162m
we also know that the force on impulse is given as
Ft=mv
F=mv/t
F=7*18/9
F=126/9
F=14N
work done = Force* distance
work done=14*162
work=2268Joules
work= 2.27kJ
can vectors be strung together?
Answer:
The head-to-tail method is a graphical way to add vectors, The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.
Explanation:
is 2/2 1 or 0? please help lol
Answer:
1.
Explanation:
Hello!
In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:
2 ÷ 2 = 1.
Best regards!
Which of the following requires the expenditure of more work?
a. Lifting a 110 newton [N] weight a height of 3 meters [m].
b. Exerting a force of 60 pounds-force [lbf] on a sofa to slide it 30 feet [ft] across a room.
Answer:
The correct answer is option B
Explanation:
Step one:
given data
a. force F= 110N
distance s= 3meters
we know that work= Force* distance
work= 110*3
Work= 330Joules
Step two:
data
Force= 60 pounds
distance= 30 ft
convert pounds to Newton
1 pound= 4.44822N
60 pounds= 60*4.44822
=266.9N
convert ft to meteres
1 ft = 0.3048meter
30ft= 0.3048*30
=9.144N
we know that work= Force* distance
work= 266.9N*9.144N
Work= 2440.53Joules
HELP ASAP!!!
Which graph shows the change in velocity of an object in free fall?
Answer:
the graph that show Change in velocity is no A
The graph that showing the velocity with respect to time for a free falling body is figure C where, the downward velocity will be higher due to the acceleration due to gravity.
What is velocity?Velocity of a moving body is the measure of the distance covered per unit time. Thus, it is the ratio of distance to the time. Velocity is expressed in the units of Km/h, m/s. miles/h, ft./s etc.
The rate of change of velocity is called acceleration. The acceleration by the force of a gravitational field is called acceleration due to gravity g having the value 9.8 m/s².
The velocity - time graph shows a diagonal relation for a free falling body. A free falling body is initially at rest and falls downward with the negative velocity with the acceleration due to gravity. Therefore, figure B shows the change in velocity of a free falling body.
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I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
a. 250 kg
c. 10 Kg
b. .002 Kg
d. 45 Kg
Answer:
a. 250kg I think it's the right answer. hope it helps:)
Answer:
C.10
Explanation:
because when you divide 50 divided by 5 = 10
A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?
Answer:
The number is [tex]N = 300[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = -4.8 *10^{-17 } \ C[/tex]
Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]
Generally the number of excess electrons is mathematically represented as
[tex]N = \frac{Q}{e}[/tex]
=> [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]
=> [tex]N = 300[/tex]
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground. On his first try, Robin looses the arrow at v0=35.0 m/sv0=35.0 m/s at an angle of θ=30.0°θ=30.0° above the horizontal. The arrow has an initial height of y0=1.50 m,y0=1.50 m, and its tip is x=60.0 mx=60.0 m away from the target orange. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has reached the horizontal position xx of the orange? Use g=9.81 m/s2g=9.81 m/s2 for the acceleration due to gravity.
Answer:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
h' = 55.3 m
At an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s. What is the angle of the shock cone
Answer:
55 degrees
Explanation:
Given that an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s.
From the question above, we can get the below parameters
Object speed (V) = 1500 km/h
Sound speed ( v) = 342 m/s
Convert km/h to m/s
(1500 × 1000)/3600
Jet speed V = 416.67 m/s
Let's first calculate the mash number M.
M = V/v
M = 416.67 / 342
M = 1.2183
Formula for the angle of the shock cone is reciprocal of mash number. That is,
Sin Ø = 1 / M
Sin Ø = 1 / 1.2183
Sin Ø = 0.8208
Ø = sin^-1(0.8208)
Ø = 55 degree
Therefore, the angle of the shock cone is approximately 55 degrees
On Earth, we experience lunar and solar eclipses. what types of eclipses (if any) would an inhabitant of the moon experience? Explain.
Answer:
However, those astronauts would experience a second spectacle: A solar eclipse caused by the Earth – the Sun disappearing behind the dark disc of the Earth. When Earth inhabitants witness a lunar eclipse, Moon inhabitants would, simultaneously be witnessing a solar eclipse.
FOR THIS QUESTION, solve as if you are on the moon and use moon gravity (g = 1.7 m/s^2)
An astronaut stands on top of a lunar lander, which is 2.4 meters tall, and putts golf balls off the side with an initial velocity of 4.5 meters per second. How far in the horizontal direction will they travel?
A.) 12.6 m
B.) 2.8 m
C.) 1.7 m
D.) 7.7 m
Answer:the answer is D
Explanation: I just took the quiz